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I want to prove that if $\sum_{n=1}^{\infty}a_n$ converges, with $a_n >0$ for all $n$, then $\sum_{n=1}^{\infty} \frac{a_n}{n}$ converges.

My book gives a proof where shows that $S_k=\sum_{n=1}^{k} \frac{a_n}{n}$ is monotonically increasing and bounded, so converges by the completeness axiom.

However can I prove it by the comparison test?

I know that $\forall n\in\mathbb{N}$ $$0\leq \frac{a_n}{n} \leq a_n$$ and also, I know that $\sum_{n=1}^{\infty}a_n$ converges, so by the comparison tests, also $\sum_{n=1}^{\infty} \frac{a_n}{n}$ converges.

Is this acceptable?

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    $\begingroup$ Absolutely correct : you are using the "comparison principle" for series with positive real terms. And the solution in your book tells exactly the same thing, except that it proves that principle in a special case. $\endgroup$ – Adren Jan 24 '17 at 9:57
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I wrote in a comment that this proof is correct.

It should be added that the stronger result holds :

Proposition If $\sum_{n\ge1}a_n$ converges, where the $a_n$ are complex, and if $\alpha$ is some positive constant, then $\sum_{n\ge1}\frac{a_n}{n^\alpha}$ converges to.

This is a consequence of the following theorem :

Theorem (Abel) If $(a_n)\in\mathbb{C}^\mathbb{N}$ and $(t_n)\in[0,+\infty[^\mathbb{N}$ are sequences such that :

  • $\exists M\ge0\,\forall n\in\mathbb{N},\,\left|\sum_{k=0}^na_k\right|\le M$

  • $\forall n\in\mathbb{N},\,t_{n+1}\le t_n$ and $\lim_{n\to\infty}t_n=0$

then the series $\sum_{n\ge0}t_na_n$ converges.

Proof If we define $A_n=\sum_{k=0}^na_k$ for all $n\in\mathbb{N}$, then (Abel's transformation) :

$$\sum_{k=1}^nt_ka_k=\sum_{k=1}^nt_k(A_k-A_{k-1})=-t_1A_0+t_nA_n+\sum_{k=1}^{n-1}(t_t-t_{k+1})A_k$$

The series $\sum_{k\ge1}(t_k-t_{k+1})A_k$ converges absolutely (and hence converges) because $\left|(t_k-t_{k+1})A_k\right|\le M(t_k-t_{k+1})$, for all $k$.

Clearly the sequence $(t_nA_n)_{n\ge0}$ converges (to zero) and this completes the proof.

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  • $\begingroup$ I've never seen this before! Do you reckon this is a typical theorem for Analysis course? $\endgroup$ – Euler_Salter Jan 24 '17 at 10:24
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    $\begingroup$ @Euler_Salter Yes, I do. An important special case of this result is the following. Consider $\theta\in\mathbb{R}$ which is not $2k\pi$ for any $k\in\mathbb{Z}$ and $\alpha>0$; then the series $\sum_{n\ge1}\frac{e^{in\theta}}{n^\alpha}$ converges. You mainly have to show that the partial sums $\sum_{k=1}^ne^{ik\theta}$ form a bounded sequence. $\endgroup$ – Adren Jan 24 '17 at 10:29

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